Cauchy-Riemann Equations

Let

$$f(x,y) \equiv u(x,y)+iv(x,y),$$ (1)

where

$$z \equiv x+iy,$$ (2)

so

$$dz = dx+i\,dy.$$ (3)

The total derivative of $f$ with respect to $z$ may then be computed as follows.

$$y = {z-x\over i}$$ (4)

$$x = z-iy,$$ (5)

so

$${\partial y\over \partial z} = {1\over i} = -i$$ (6)

$${\partial x\over \partial z} = 1,$$ (7)

and

$${df\over dz} = {\partial f\over \partial x} {\partial x\over... ... {\partial f\over \partial x} -i {\partial f\over \partial y}.$$ (8)

In terms of $u$ and $v$, (8) becomes

$${df\over dz} = \left({{\partial u\over\partial x}+i{\partial v\over\partial x}}\right)- i\left({{\partial u\over\partial y}+i{\partial v\over\partial y}}\right)$$

$$= \left({{\partial u\over \partial x}+ i {\partial v\over\partial x... ...t)+ \left({-i{\partial u\over \partial y}+ {\partial v\over\partial y}}\right).$$ (9)

Along the real, or x-Axis, ${\partial f/\partial y} = 0$, so

$${df\over dz} = {\partial u\over \partial x} + i {\partial v\over\partial x}.$$ (10)

Along the imaginary, or $y$-axis, ${\partial f/\partial x} = 0$, so

$${df\over dz} = - i {\partial u\over \partial y} + {\partial v\over \partial y}.$$ (11)

If $f$ is Complex Differentiable, then the value of the derivative must be the same for a given $dz$, regardless of its orientation. Therefore, (10) must equal (11), which requires that

$${\partial u\over\partial x} = {\partial v\over\partial y}$$ (12)

and

$${\partial v\over \partial x} = - {\partial u\over \partial y}.$$ (13)

These are known as the Cauchy-Riemann equations. They lead to the condition

$${\partial^2u\over \partial x\partial y} = - {\partial^2v\over \partial x\partial y}.$$ (14)

The Cauchy-Riemann equations may be concisely written as

$${df\over dz^*} = {\partial f\over \partial x} +i {\partial f\over \partial y} = \l... ...)+i\left({{\partial u\over \partial y} + i {\partial v\over \partial y}}\right)$$

$$= \left({{\partial u\over \partial x} - {\partial v\over \partial y... ...i\left({{\partial u\over \partial y} + {\partial v\over \partial x}}\right)= 0.$$ (15)

In Polar Coordinates,

$$f(re^{i\theta})\equiv R(r,\theta)e^{i\Theta(r,\theta)},$$ (16)

so the Cauchy-Riemann equations become

$${\partial R\over \partial r} = {R\over r} {\partial \Theta \over \partial \theta }$$ (17)

$${1\over r} {\partial R\over \partial \theta } = -R {\partial \Theta \over \partial r}.$$ (18)

If $u$ and $v$ satisfy the Cauchy-Riemann equations, they also satisfy Laplace's Equation in 2-D, since

$${\partial^2 u\over \partial x^2}+{\partial^2 u\over \partial... ...\over\partial y}\left({-{\partial v\over\partial x}}\right)= 0$$ (19)

$${\partial^2 v\over \partial x^2}+{\partial^2 v\over \partial... ...al\over\partial y}\left({\partial u\over\partial x}\right)= 0.$$ (20)

By picking an arbitrary $f(z)$, solutions can be found which automatically satisfy the Cauchy-Riemann equations and Laplace's Equation. This fact is used to find so-called Conformal Solutions to physical problems involving scalar potentials such as fluid flow and electrostatics.

See also Cauchy Integral Theorem, Conformal Solution, Monogenic Function, Polygenic Function

References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 17, 1972.

Arfken, G. ``Cauchy-Riemann Conditions.'' §6.2 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 3560-365, 1985.